A normally distributed population has a mean of 118 with a standard deviation of 18. What score separates the lowest 72% of the distribution from the rest of the scores?

To find the score that separates the lowest 72% of the distribution from the rest of the scores, we need to find the z-score corresponding to the 72nd percentile and then convert it back to the original score. First, we find the z-score using the standard normal distribution table or calculator. The 72nd percentile corresponds to a z-score of approximately 0.6052. Next, we use the z-score formula to convert the z-score back to the original score: Z = (X - μ) / σ Rearranging the formula to solve for X, we have: X = Z * σ + μ Substituting the values, we get: X = 0.6052 * 18 + 118 X ≈ 128.8936 Therefore, the score that separates the lowest 72% of the distribution from the rest of the scores is approximately 128.8936.